Does simultaneous measurement imply that we can only use $1$ copy of quantum state to measure any set of commuting observables? For example, suppose we have a Bell state $(\lvert 00 \rangle + \lvert 11 \rangle) / \sqrt{2} $, and want to measure this state to $X \otimes X$ and then $Z \otimes Z$ observables. Then, measuring $X \otimes X$ wouldn't change the state so that we can measure $X \otimes X$ and then $Z \otimes Z$ so that we are essentially re-using the state and thus the total number of copy we used is just $1$. Is this a correct understanding of a simultaneous measurement, and does this hold in general when the operators commute?
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Yes, that's actually precisely the reason why simultaneous measurement is so useful, and any mutually commuting set of observables is simultaneously measurable.
